Question

In an AC circuit containing only an inductor, the current

• A. figure A
• B. figure B
• C. figure 3
• D. figure 4

Hint:

Remember that for a purely inductive circuit, the phase difference between voltage and current is always $\frac{\pi}{2}$ radians or 90 degrees.

Answer 1

The Correct Option is A

Step 1: Set up the circuit.
The circuit has only an inductor connected to an alternating voltage. We want to know how the current behaves compared to the voltage.

Step 2: Apply the inductor rule.
The voltage across an inductor depends on how fast the current changes. \[ V = L\frac{dI}{dt} \]

Step 3: Use a sine voltage.
Let the source voltage be \[ V(t) = V_0\sin(\omega t) \] We solve for the current that produces this voltage.

Step 4: Find the current.
Working out the current that fits the equation gives \[ I(t) = \frac{V_0}{\omega L}\sin\left(\omega t - \frac{\pi}{2}\right) \]

Step 5: Read the phase.
The current has the same shape as the voltage but is shifted by $\dfrac{\pi}{2}$, or ninety degrees, behind it. So the current lags the voltage.

Step 6: State the meaning.
In a pure inductor the current lags the voltage by a quarter cycle, which is the graph shown in option A. \[ \boxed{\text{Current lags voltage by } \frac{\pi}{2}} \]